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Course: AERO 310, Spring 2012
School: Texas A&M
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325 A 123 Question particle of mass m slides without friction along a track in the form of a parabola as shown in Fig. P3-25. The equation for the parabola is y= r2 2a where a is a constant, r is the distance from point O to point Q, point Q is the projection of point P onto the horizontal direction, and y is the vertical distance. Furthermore, the particle is attached to a linear spring with spring constant K...

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325 A 123 Question particle of mass m slides without friction along a track in the form of a parabola as shown in Fig. P3-25. The equation for the parabola is y= r2 2a where a is a constant, r is the distance from point O to point Q, point Q is the projection of point P onto the horizontal direction, and y is the vertical distance. Furthermore, the particle is attached to a linear spring with spring constant K and unstretched length x0 . The spring is always aligned horizontally such that its attachment point is free to slide along a vertical shaft through the center of the parabola. Knowing that the parabola rotates with constant angular velocity (where = ) about the vertical direction and that gravity acts vertically downward, determine the dierential equation of motion for the particle in terms of the variable r . g m Q y O r P Figure P 3-25 Solution to Question 325 Kinematics For this problem it is convenient to dene a xed inertial reference frame F and a non-inertial reference frame A. Corresponding to reference frame F , we choose the following coordinate system: Ex Ey Ez Origin at Point O = = = Along OP When t = 0 Along Oy When t = 0 Ex Ey 124 Chapter 3. Kinetics of Particles Furthermore, corresponding to reference frame A, we choose the following coordinate system: Origin at Point O ex = Along OP ey = Along Oy ez = ex ey The position of the particle is then given in terms of the basis {ex , ey , ez } as r = x ex + y ey = x ex + (x 2 /a)ey (3.554) Furthermore, since the parabola spins about the ey -direction, the angular velocity of reference frame A in reference frame F is given as F A = = ey (3.555) The velocity in reference frame F is then found using the rate of change transport theorem as F dr Adr F A F v= = + r (3.556) dt dt Using r from Eq. (3.554) and F A from Eq. (3.555), we have that A dr = x ex + (2x x/a)ey dt FA r = ey (x ex + (x 2 /a)ey ) = x ez (3.557) (3.558) Adding the expressions in Eq. (3.557) and Eq. (3.558), we obtain F v as F v = x ex + (2x x/a)ey x ez (3.559) The acceleration in reference frame F is found by applying the rate of change transport theorem to F v as A F dF dF F v= v + F A F v (3.560) a= dt dt Using F v from Eq. (3.559) and F A from Eq. (3.555), we have that A dF v = x ex + 2(x 2 + x x)/a ey x ez (3.561) dt FA F v = ey (x ex + (2x x/a)ey x ez ) = x ez 2 x ex (3.562) Adding the expressions in Eq. (3.561) and Eq. (3.562) and, we obtain F a as F a = (x 2 x)ex + 2(x 2 + x x)/a ey 2x ez (3.563) 125 Nn Fs Nb mg Figure 3-22 Free Body Diagram of Particle for Question 325. Kinetics The free body diagram of the particle is shown in Fig. 3-22. Using Fig. 3-22, it is seen that the following forces act on the particle: Nn = Reaction Force of Track on Particle Normal to Track and In Plane of Parabola Nb = Reaction Force of Track on Particle Normal to Track and Orthogonal to Plane of Parabola Fs = Spring Force mg = Force of Gravity Given the description of the two reaction forces Nn and Nb , we have that Nn = Nn en (3.564) Nb = Nb eb (3.565) where en and eb are the principle unit normal and principle unit vector bi-normal to the parabola. Now since reference frame A is the reference frame of the parabola, the tangent vector to the parabola is given as Av et = Now we have Av (3.566) Av from Eq. (3.557) as A v = x ex + (2x x/a)ey (3.567) Consequently, et = x ex + (2x x/a)ey x 2x 1+ a 2 = ex + (2x/a)ey 2x 1+ a 2 (3.568) Next, we know that eb must lie orthogonal to the plane of the parabola. Consequently, we have that eb = ez (3.569) 126 Chapter 3. Kinetics of Particles Therefore, en = eb et = ez ex + (2x/a)ey 2x 1+ a 2 = (2x/a)ex + ey 2x 1+ a 2 (3.570) Suppose now that we dene = Then we can write en = 2x a 1+ 2 (2x/a)ex + ey (3.571) (3.572) The reaction force exerted by the parabola on the particle is then given as N = Nn + Nb = Nn (2x/a)ex + ey + Nb ez (3.573) Next, the spring force is given as Fs = K( 0 )us Now for this problem we know that the unstretched length of the spring is Furthermore, the stretched length of the spring is given as = r rQ (3.574) 0 = x0 . (3.575) where P is the attachment point of the spring. Now since the attachment point lies on the ey -axis at the same value of y as the particle, we have that rQ = y ey = x2 ey a (3.576) Therefore, r rQ = x ex + x2 x2 ey ey = x ex a a (3.577) The stretched length of the spring is then given as = x ex = x (3.578) Finally, the direction of the spring force is given as us = r rQ r rQ = x ex = ex x (3.579) The force of the linear spring is then given as Fs = K(x x0 )ex (3.580) 127 Finally, the force of gravity is given as mg = mg ey (3.581) Then, adding Eq. (3.573), Eq. (3.580), and Eq. (3.581), the resultant force acting on the particle is then obtained as F = Nn + Nb + Fs + mg = Nn (2x/a)ex + ey + Nb ez K(x x0 )ex mg ey (3.582) Then, combining terms with common components in Eq. (3.582), we have that F = Nn 2x/a + K(x x0 ) ex + Nn mg ey + Nb ez (3.583) Then, setting F from Eq. (3.583) equal to mF a using the expression for F a from Eq. (3.563), we obtain Nn 2x/a + K(x x0 ) ex + Nn mg ey + Nb ez = m(x 2 x)ex + m 2(x 2 + x x)/a ey 2mx ez (3.584) Equating components in Eq. (3.584), we obtain the following three scalar equations: m(x 2 x) = Nn m 2(x 2 + x x)/a 2 m x 2x/a K(x x0 ) Nn mg = Nb = (3.585) (3.586) (3.587) Now the dierential equation of motion for the particle is obtained as follows. Rearranging Eq. (3.585) and Eq. (3.586), we have that Nn 2x/a Nn = m(x 2 x) K(x x0 ) (3.588) = m 2(x 2 + x x)/a + mg (3.589) Then, dividing Eq. (3.588) by Eq. (3.589), we obtain Nn 2x/a m(x 2 x) K(x x0 ) = Nn m 2(x 2 + x x)/a + mg (3.590) 128 Chapter 3. Kinetics of Particles Simplifying Eq. (3.590) gives 2x m(x 2 x) K(x x0 ) = a m 2(x 2 + x x)/a + mg (3.591) Multiplying Eq. (3.591) by m 2(x 2 + x x)/a + mg , we obtain m 2(x 2 + x x)/a + mg 2x = m(x 2 x) K(x x0 ) a (3.592) Rearranging Eq. (3.592), we have that m(x 2 x) + K(x x0 ) + m 2(x 2 + x x)/a + mg 2x =0 a (3.593) Simplifying further, we obtain the dierential equation of motion as mx 1 + 2x a 2 +m 2g x 2 x + K(x x0 ) + 4mx a a 2 =0 (3.594)
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